MHB Free Modules - Another problem regarding Bland Proposition 2.2.3

  • Thread starter Thread starter Math Amateur
  • Start date Start date
  • Tags Tags
    Modules
Math Amateur
Gold Member
MHB
Messages
3,920
Reaction score
48
I am reading Paul E. Bland's book, "Rings and Their Modules".

I am trying to understand Section 2.2 on free modules and need help with the proof of Proposition 2.2.3 - that is some further help ... (for my first problem with the proof see my post "http://mathhelpboards.com/linear-abstract-algebra-14/free-modules-bland-proposition-2-2-3-a-13179.html")

Proposition 2.2.3 and its proof read as follows:View attachment 3531The first line of Bland's proof reads as follows:

"Proof: $$(1) \Longrightarrow (2)$$. Since $$ \{ x_\alpha \}_\Delta$$ is a set of generators of $$M $$, it is certainly the case that each $$x \in M$$ can be expressed as $$x = \sum_\Delta x_\alpha a_\alpha$$ where $$a_\alpha = 0$$ for almost all $$\alpha \in \Delta$$. ... ... "Now, by definition of a basis, $$ \{ x_\alpha \}_\Delta$$ being a basis for M certainly implies that each $$x \in M$$ can be expressed as a sum $$x = \sum_\Delta x_\alpha a_\alpha$$ ... ...

... ... BUT ... ... how do we know that each $$x \in M$$ can be expressed as $$x = \sum_\Delta x_\alpha a_\alpha$$ where $$a_\alpha = 0$$ for almost all $$\alpha \in \Delta$$ ... ...

... ... specifically ... ... what/where in the definition of a basis is the justification for adding the condition "where $$a_\alpha = 0$$ for almost all $$\alpha \in \Delta$$"?

Hope someone can help,

Peter
 
Physics news on Phys.org
Peter said:
I am reading Paul E. Bland's book, "Rings and Their Modules".

I am trying to understand Section 2.2 on free modules and need help with the proof of Proposition 2.2.3 - that is some further help ... (for my first problem with the proof see my post "http://mathhelpboards.com/linear-abstract-algebra-14/free-modules-bland-proposition-2-2-3-a-13179.html")

Proposition 2.2.3 and its proof read as follows:View attachment 3531The first line of Bland's proof reads as follows:

"Proof: $$(1) \Longrightarrow (2)$$. Since $$ \{ x_\alpha \}_\Delta$$ is a set of generators of $$M $$, it is certainly the case that each $$x \in M$$ can be expressed as $$x = \sum_\Delta x_\alpha a_\alpha$$ where $$a_\alpha = 0$$ for almost all $$\alpha \in \Delta$$. ... ... "Now, by definition of a basis, $$ \{ x_\alpha \}_\Delta$$ being a basis for M certainly implies that each $$x \in M$$ can be expressed as a sum $$x = \sum_\Delta x_\alpha a_\alpha$$ ... ...

... ... BUT ... ... how do we know that each $$x \in M$$ can be expressed as $$x = \sum_\Delta x_\alpha a_\alpha$$ where $$a_\alpha = 0$$ for almost all $$\alpha \in \Delta$$ ... ...

... ... specifically ... ... what/where in the definition of a basis is the justification for adding the condition "where $$a_\alpha = 0$$ for almost all $$\alpha \in \Delta$$"?

Hope someone can help,

Peter

Let $M$ be any $R$-module and $S$ be a subset of $M$.

Then the submodule of $M$ generated by $S$ is defined as

$$\{ \sum_{\text{finite}}a_is_i:a_i\in R, s_i\in S \}$$

Summing up infinitely many elements of a module doesn't make any sense.

Now coming to your problem.
If $X=\{x_\alpha\}_{\alpha\in J}$ is a basis of $M$, then $X$ generates $M$. Use the definition of "generation" as discussed above.

I hope this helped.
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top